metode eliminasi gauss

Invers Matriks 

Cara menghitung invers matriks berordo 3x3 ada dua cara, yaitu dengan metode Adjoint, dan metode OBE (Operasi Baris Elementer). Pada saat ini saya akan menghitung invers menggunakan metode Adjoint dengan bahasa pemrograman Python3, 

berikut source code-nya / program nya :

import os

a=1
while a==1:
    print("===Angka matriks===")
    a=int(input("Nilai M11 = "))
    b=int(input("Nilai M12 = "))
    c=int(input("Nilai M13 = "))
    d=int(input("Nilai M14 = "))
    e=int(input("Nilai M15 = "))
    f=int(input("Nilai M16 = "))
    g=int(input("Nilai M17 = "))
    h=int(input("Nilai M18 = "))
    i=int(input("Nilai M19 = "))

#Determinan
    detA=(a*e*i)+(b*f*g)+(c*d*h)-(g*e*c)-(h*f*a)-(i*d*b)
    print("\nMATRIKS ")
    print("|`",a," ",b," ",c,"`|")
    print("| ",d," ",e," ",f," |  ===> DetA = ",detA)
    print("|_",g," ",h," ",i,"_|")
    print("")
#Adjoin
    a11=(e*i)-(h*f)
    a12=(d*i)-(g*f)
    a13=(d*h)-(g*e)
    a21=(b*i)-(h*c)
    a22=(a*i)-(g*c)
    a23=(a*h)-(g*b)
    a31=(b*f)-(e*c)
    a32=(a*f)-(d*c)
    a33=(a*e)-(d*b)
    print('ADJOIN')
    print("A11 = (+)[ ",e,"",f," ]=",a11," A12 = (-)[ ",d,"",f," ]=",a12," A13 = (+)[ ",d,"",e," ]=",a13)
    print("         [ ",h,"",i," ]                  [ ",g,"",i," ]                  [ ",g,"",h," ]")

    print("\nA21 = (-)[ ",b,"",c," ]=",a21," A22 = (+)[ ",a,"",c," ]=",a22," A23 = (-)[ ",a,"",b," ]=",a23)
    print("           [ ",h,"",i," ]                  [ ",g,"",i," ]                  [ ",g,"",h," ]")

    print("\nA31 = (+)[ ",b,"",c," ]=",a31," A32 = (-)[ ",a,"",c," ]=",a32," A33 = (+)[ ",a,"",b," ]=",a33)
    print("           [ ",e,"",f," ]                  [ ",d,"",f," ]                  [ ",d,"",e," ]")

#printAdjoin
    print("Adj = [ ",a11*(1),"",a12*(-1),"",a13*(1)," ]")
    print("      [ ",a21*(-1),"",a22*(1),"",a23*(-1)," ]")
    print("      [ ",a31*(1),"",a32*(-1),"",a33*(1)," ]")
    print("")
    print("A Traspose = [ ",a11*(1),'',a21*(-1),'',a31*(1),' ]')
    print('             [ ',a12*(-1),'',a22*(1),'',a32*(-1),' ]')
    print('             [ ',a13*(1),'',a23*(-1),'',a33*(1),' ]')

#invers
    INVERS_11=(1/detA*(a11*(1)))
    INVERS_12=(1/detA*(a12*(-1)))
    INVERS_13=(1/detA*(a13*(1)))
    INVERS_21=(1/detA*(a21*(-1)))
    INVERS_22=(1/detA*(a22*(1)))
    INVERS_23=(1/detA*(a23*(-1)))
    INVERS_31=(1/detA*(a31*(1)))
    INVERS_32=(1/detA*(a32*(-1)))
    INVERS_33=(1/detA*(a33*(1)))

    print("\nA-1 = 1/Det A x Adj A")
    print('    = 1/',detA,"[ ",a11,'',a21,'',a31," ]")
    print('                [ ',a12,'',a22,'',a32,' ]')
    print('                [ ',a13,'',a23,'',a33,' ]')

    print('\n    =  [ ',INVERS_11,'',INVERS_21,'',INVERS_31,' ]')
    print('        [ ',INVERS_12,'',INVERS_22,'',INVERS_32,' ]')
    print('        [ ',INVERS_13,'',INVERS_23,'',INVERS_33,' ]')

    input("\nRismonita anggraeni")
    os.system('cls')
 

 Demikian  ini lah cara menghitung invers matriks berordo 3x3 dengan metode Adjoint menggunkan bahasa pemrograman Python3. 

Terima kasih dan semoga bermanfaat.